Question

What is the molecular weight of acetic acid if a solution that contains 30.0 grams of acetic acid per kilogram of water freezes at$-0.93{}^\circ C$. Do these results agree with the assumption that acetic acid has the formula$C{{H}_{3}}C{{O}_{2}}H$ ?

Answer 1

Acetic acid is an organic weak acid. The depression in freezing point is directly proportional to the molality of any compound. The molality is the moles of solute per kg of solvent. As the moles involve mass and molar mass, the molecular formula of acetic acid can be determined.

Complete answer:
We have to find the molecular weight of acetic acid from the data given, that is the mass of acetic acid is 30.0 g/ kg water and its freezing point is $-0.93{}^\circ C$. We have to compare and check this calculated value with the original molar mass value of acetic acid, which is $C{{H}_{3}}C{{O}_{2}}H$.
For this we will calculate the molality and then molar mass from molality. As we know, from depression in freezing point that, $\Delta {{T}_{f}}={{K}_{f}}.m$ , where ${{K}_{f}}$, molal depression constant for water is $1.86{}^\circ C\,kg\,mo{{l}^{-1}}$ , and the depression in freezing point$\Delta {{T}_{f}}$ is the difference of final and initial freezing point,${{T}_{f}}^{0}-{{T}_{f}}$ = $0.00{}^\circ C-(-0.93){}^\circ C$ = $0.93{}^\circ C$. Now solving for molality (m), we have,
$m=\dfrac{\Delta {{T}_{f}}}{{{K}_{f}}}$
$m=\dfrac{0.93{}^\circ C}{1.86{}^\circ C.kg.mo{{l}^{-1}}}$
m = 0.50 moles/kg
From this molality we will find molar mass, as we know, from the formula of number of moles, $no.\,of\,moles=\dfrac{given\,mass}{molar\,mass}$ , rearranging this to find the molar mass, we have,
Molar mass = $\dfrac{30.0\,g}{0.50\,mol}$
Molar mass = 60 g/mol
So, the molar mass of acetic acid is 60 u.
Now, from the given formula of acetic acid $C{{H}_{3}}C{{O}_{2}}H$, we will calculate the molar mass as,
Molar mass of $C{{H}_{3}}C{{O}_{2}}H$= mass of $\left( C+3\times H+C+2\times O+H \right)$
Putting the masses of carbon, hydrogen and oxygen we have,
Molar mass of $C{{H}_{3}}C{{O}_{2}}H$= $\left( 12+3\times 1.008+12+2\times 16+1.008 \right)$
Molar mass of $C{{H}_{3}}C{{O}_{2}}H$= 60.03 u
Hence, we get the calculated value of acetic acid to be 60 u that agrees with the assumptions from the molecular mass of $C{{H}_{3}}C{{O}_{2}}H$ to be 60.03 u.

Note:
${{K}_{f}}$ is the molal depression or cryoscopic constant, which has different values for different solvents. We have ${{T}_{f}}^{0}-{{T}_{f}}$where ${{T}_{f}}^{0}$is the initial freezing point and ${{T}_{f}}$ as final freezing point. Also Van’t hoff factor ‘i’ is applied in the freezing point depression formula, as $\Delta {{T}_{f}}=i.{{K}_{f}}.m$, here it will have a value of 1.